如果是元音则如何删除第一个字母，如果没有元音则如何不返回元音

Question

Write the function startWithVowel(word) that takes in a word as argument and returns a substring that starts with the first vowel found in the word. The function returns 'No vowel' if the word does not contain vowel.

Examples

>>> startWithVowel('apple')
'apple'
>>> startWithVowel('google')
'oogle'
>>> startWithVowel('xyz')
'No vowel'

My Answer

def startWithVowel(word):
    if word[0] in 'aeiou':
        return word
    elif word[0] != 'aeiou' and word[1:] in 'aeiou':
        return word[1::]
    elif word not in 'aeiou':
        return "No vowel"

MyQuestion I know how to remove first vowel letter if word contains any vowel, but i am stuck with this code.. using this logic it should return 'no vowel' for xyz. but it's returning 'yz' and i know where i am wrong, it's a logical problem in line 4. but i don't know how to solve this problem.

Answer 1

Here is a rather straightforward way to do it:

def startWithVowel(word):
    for i, c in enumerate(word):
        if c.lower() in 'aeiou':
            return word[i:]
    return 'No vowel'

>>> for word in 'apple', 'google', 'xyz', 'BOGGLE':
...     print startWithVowel(word)
apple
oogle
No vowel
OGGLE

Answer 2

How about this -

>>> def startWithVowel(word):
...     while len(word) > 0 and word[0] not in 'aeiou':
...             word = word[1:]
...     if len(word) == 0:
...             return 'No vowel'
...     return word
...
>>> startWithVowel('apple')
'apple'
>>> startWithVowel('google')
'oogle'
>>> startWithVowel('xyz')
'No vowel'
>>> startWithVowel('flight')
'ight'

Use word[0].lower() in the above loop, if you want to check case-insensitive.

Answer 3

The problem with your logic in line 4 is that you are using != instead of not in . What you are doing is comparing word[0] which is a character and comparing it to the string 'aeiou' . Naturally a single character will never be equal to a string. On top of that, word[1:] in 'aeiou' and word in 'aeiou' don't work. They are comparing strings, not iterating over each letter of the word and comparing the characters. You can fix this by doing the following

def startWithVowel(word):
    if word[0] in 'aeiou':
        return word
    else:
        for i in range(1, len(word)):
            if word[i] in 'aeiou': return word[i:]
        return "No vowel"

This says if the first letter is a vowel return the word, else, iterate over each letter in the word and look for a vowel. If a vowel is found then return the word from that index and if not then return 'No vowel'

Answer 4

You may use re.search .

import re
def findstring(s):
    m = re.search(r'(?i)[aeiou].*', s)
    if m:
        return m.group()
    else:
        return "No vowel"

Answer 5

def startWithVowel(word):
dataLen=len(word)
newStr=''
if word[0] in 'aieou':
   return word
for xx in range(1,dataLen):
     count=0
     if word[0] not in 'aieou' and word[xx] in "aieou" or word[xx] not in 'aieou':
        newStr=newStr+word[xx]
        if(word[xx] in 'aieou'):
           count=count+1
if count>0:
   return newStr
else:
    return 'No vowel'          
#return newStr
for xx in word:
    if xx not in 'aieou':
       return 'No vowel'